Question

An element belongs to period 4 and its outermost shell contains 2 electrons. The last electron is added to its penultimate shell. The four sets of quantum numbers possible for that electron present in the penultimate shell is

• A. n=4,l=1,m=3,s=+1/2
• B. n=4,l=3,m=7,=-1/2
• C. n=3,l=2,m=5,s=+1/2
• D. n=3,l=1,m=3,s-1/2

Answer 1

Answer:

C. n=3,l=2,m=5,s=+1/2

• Principal Quantum Number (n): Indicates the energy level or shell of an electron. For the given element, the outermost shell is the 4th period, meaning n=4.

• Azimuthal Quantum Number (l): Describes the shape of the orbital and the subshell. It ranges from 0 to n-1. l=0 for s-subshell, l=1 for p-subshell, l=2 for d-subshell, and l=3 for f-subshell.

• Magnetic Quantum Number (m): Specifies the orientation of the orbital in space. It ranges from -l to +l, including 0.

• Spin Quantum Number (s): Represents the intrinsic angular momentum of an electron, which can be either spin up (+1/2) or spin down (-1/2).

Interpreting the Problem

• The element is in Period 4, so its valence shell is n=4.

• The outermost shell contains 2 electrons. This suggests that the element is an alkaline earth metal (Group 2), and its electronic configuration ends in 4s².

• The crucial piece of information is that the last electron is added to its penultimate shell. The penultimate shell is the shell just before the outermost shell. Since the outermost shell is n=4, the penultimate shell is n=3.

• This implies that the 2 electrons in the outermost shell (4s²) were added after the penultimate shell was being filled or was already filled. For an element in Period 4 with 2 electrons in the outermost shell, the filling order follows the Aufbau principle. After filling up to 3p⁶, the next electrons go into 4s. Thus, the penultimate shell is n=3.